YES

Problem:
 a(x1) -> x1
 a(b(x1)) -> c(b(x1))
 a(c(c(x1))) -> c(c(a(a(x1))))

Proof:
 DP Processor:
  DPs:
   a#(c(c(x1))) -> a#(x1)
   a#(c(c(x1))) -> a#(a(x1))
  TRS:
   a(x1) -> x1
   a(b(x1)) -> c(b(x1))
   a(c(c(x1))) -> c(c(a(a(x1))))
  Arctic Interpretation Processor:
   dimension: 2
   usable rules:
    a(x1) -> x1
    a(b(x1)) -> c(b(x1))
    a(c(c(x1))) -> c(c(a(a(x1))))
   interpretation:
              [0 0]     [-&]
    [b](x0) = [2 2]x0 + [1 ],
    
    [a#](x0) = [0 2]x0,
    
              [0  0 ]     [0]
    [a](x0) = [-& 0 ]x0 + [1],
    
              [-& 0 ]     [1]
    [c](x0) = [2  0 ]x0 + [0]
   orientation:
    a#(c(c(x1))) = [4 4]x1 + [5] >= [0 2]x1 = a#(x1)
    
    a#(c(c(x1))) = [4 4]x1 + [5] >= [0 2]x1 + [3] = a#(a(x1))
    
            [0  0 ]     [0]           
    a(x1) = [-& 0 ]x1 + [1] >= x1 = x1
    
               [2 2]     [1]    [2 2]     [1]           
    a(b(x1)) = [2 2]x1 + [1] >= [2 2]x1 + [1] = c(b(x1))
    
                  [2 2]     [3]    [2 2]     [3]                 
    a(c(c(x1))) = [2 2]x1 + [3] >= [2 2]x1 + [3] = c(c(a(a(x1))))
   problem:
    DPs:
     
    TRS:
     a(x1) -> x1
     a(b(x1)) -> c(b(x1))
     a(c(c(x1))) -> c(c(a(a(x1))))
   Qed